Relationship between PN conduction voltage drop and current and temperature

captz

Relationship between PN conduction voltage drop and current and temperature [Copy link]

*) , the E junction is affected by temperature, and the change in on-state voltage drop is related to Is and Ic

The conduction voltage drop of the PN junction changes inversely with temperature. It is generally believed that the temperature drops by 2mV for every degree of temperature rise . According to simulation tests, at a temperature of 27°C , transistor parameters Is=1e-13 , β=100 , β-=1 , and Ic=1mA , the change in the E junction voltage drop per unit temperature is -2mV/°C (the "-" sign indicates the opposite change ) ; and for every 10 times reduction in Is , it decreases by 0.2 mV/°C , and the temperature variable t , the E junction voltage drop decrease Vbe=-t[2-0.2(N-13)] (N is the exponent of Is ) . For example, Is=1e-18 exponent N=18 , Vbe=-t(2-0.2*5)=-t(1)(mV) . The simulation verification is as follows: Is=1e-16 , that is, Vbe=-t(2-0.2*3)=102.2mV .

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Learned

captz

The E junction voltage drop decreases by 3mV as the unit collector current (mA) increases , that is -3mV/(Ic-1)mA . Therefore, when the temperature changes, the E junction voltage drop variable calculation is modified to Vbe=-t[2-0.2(N-13)]+3(Ic-1)(mV) . For example, in the following circuit, Ic=3.7mA , that is, Vbe=-102.2+3(3.7-1)=-94.1mV . The simulation verification is as follows.

freebsder

Good details, learned a lot.

captz

captz posted on 2021-8-25 16:44 The E junction voltage drop decreases by 3mV as the unit collector current (mA) increases, that is -3mV/(Ic-1)mA. Therefore, when the temperature changes, the E junction voltage drop variable calculation is modified to W ...

*) , temperature rise, base current increment Ibe = Vbe/Rb , substitute Rb = (Vcc-Vbe)/Ibe , Vbe = VtLn (Ie/Is) and get, Ibe = IbeVbe/[Vcc-VtLn (Ie/Is)]

For example, in the above two figures, Ibe=1000(10e-6)(102.2e-3)/[6000-VtLn(1.01e+13)]=1.956e-7=0.1956uA ( where " 1000 " is the Rb unit K)

ΔIbe=1000(37e-6)(94.1e-3)/[6000-VtLn(3.7e+13)]=0.6706uA

captz

captz posted on 2021-8-25 16:53 *), temperature rise, base current increment Ibe=Vbe/Rb, substitute Rb=(Vcc-Vbe)/Ibe, Vbe=VtLn(Ie/Is) to get...

If the amplification parameter β ≠ 100 , first calculate Ibe' according to the formula , and then substitute the actual β into Ibe=Ibe'100/β to calculate. For example, Ibe'=0.1956uA , Ibe=0.1956*100/70=0.279uA as shown in the figure below.

qwqwqw2088

How are the temperature values in the figure obtained? Are they simulated using simulation software?

captz

qwqwqw2088 posted on 2021-8-25 17:16 How are the temperature values in the figure obtained? Is it simulated by simulation software?

Set the temperature in the following fields

captz

captz posted on 2021-8-25 17:28 Set the temperature in the following column

With the above quantitative analysis, the following circuit parameters of temperature compensation of the transistor operating point can be calculated.

*) , using PN junction temperature compensation

The change in base current with temperature, Ibe , must be shunted off to keep Ibe constant. A diode with the same Is parameter is directly connected in parallel to the E junction, Id=Ie=IsExp(Vbe/Vt) ; when the temperature changes, the current variable Id ≈ Ic= Ibe . Obviously, the base current is inconsistent with the change in the diode conduction current, and appropriate temperature compensation cannot be achieved.

captz

captz posted on 2021-8-26 18:31 With the above quantitative analysis, the following circuit parameters of temperature compensation of the transistor operating point can be calculated. *), using PN junction temperature compensation...

The series resistor Rd limits the diode conduction current to 1/10 of the base current and its increment , that is, Id=0.1(Ie/β+2 Ibe) , which is suitable for compensation. Ibe remains unchanged after the temperature rises. As shown in the figure below,

Temperature rise 73°C , base current increment Ibe=1000(10e-6)(102.2e-3)/[6000-VtLn(1.01e+13)]=0.1956uA , that is, Id=0.1(1010/100+2*0.1956)=1.05uA , input current,

Ii=Ib+Id=11.05uA , Vbe=VtLn[(1e-3)/(1e-16)]=774.4mV , base resistance,

Rb=(Vcc-Vbe)/Ii=(6000-774.4)/11.05=472.9K , and the diode voltage drop,

Vd=VtLn(1.05e+10)=596.9mV , series resistance,

Rd=(774.4-596.9)/1.05=169K , build a circuit simulation, the temperature rises from 27°C to 100°C , and Ic remains unchanged.